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A Level Chemistry B (Salters) H433/02 Scientific literacy in chemistry Sample Advance Notice Article Version 1.1
For issue on or after: Date/Year
* 0 0 0 0 0 0 *
NOTES FOR GUIDANCE (CANDIDATES) 1. This leaflet contains an article which is needed in preparation for a question in the
externally assessed examination H433/02 Scientific literacy in chemistry. 2. You will need to read the article carefully and also have covered the learning
outcomes for A Level in Chemistry B (Salters). The examination paper will contain questions on the article. You will be expected to apply your knowledge and understanding of the work covered in A Level in Chemistry B (Salters) to answer this question. There are 20–25 marks available on the question paper for this question.
3. You can seek advice from your teacher about the content of the article and you can discuss it with others in your class. You may also investigate the topic yourself using any resources available to you.
4. You will not be able to bring your copy of the article, or other materials, into the examination. The examination paper will contain a fresh copy of the article as an insert.
5. You will not have time to read this article for the first time in the examination if you are to complete the examination paper within the specified time. However, you should refer to the article when answering the questions.
This document consists of 8 pages. Any blank pages are indicated.
INSTRUCTION TO EXAMS OFFICER/INVIGILATOR
Do not send this Insert for marking; it should be retained in the centre or recycled. Please contact OCR Copyright should you wish to re-use this document.
a. where a candidate crosses out an answer and provides an alternative response, the crossed out response is not marked and gains no
marks
b. if a candidate crosses out an answer to a whole question and makes no second attempt, and if the inclusion of the answer does not
cause a rubric infringement, the assessor should attempt to mark the crossed out answer and award marks appropriately.
6. Always check the pages (and additional objects if present) at the end of the response in case any answers have been continued there. If the
candidate has continued an answer there then add a tick to confirm that the work has been seen.
7. There is a NR (No Response) option. Award NR (No Response)
- if there is nothing written at all in the answer space
- OR if there is a comment which does not in any way relate to the question (e.g. ‘can’t do’, ‘don’t know’)
- OR if there is a mark (e.g. a dash, a question mark) which isn’t an attempt at the question.
Note: Award 0 marks – for an attempt that earns no credit (including copying out the question).
8. The scoris comments box is used by your Team Leader to explain the marking of the practice responses. Please refer to these comments
when checking your practice responses. Do not use the comments box for any other reason.
If you have any questions or comments for your Team Leader, use the phone, the scoris messaging system, or email.
9. Assistant Examiners will send a brief report on the performance of candidates to their Team Leader (Supervisor) via email by the end of the
marking period. The report should contain notes on particular strengths displayed as well as common errors or weaknesses. Constructive
criticism of the question paper/mark scheme is also appreciated.
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10. For answers marked by levels of response: Read through the whole answer from start to finish, concentrating on features that make it a stronger or weaker answer using the indicative scientific content as guidance. The indicative scientific content indicates the expected parameters for candidates’ answers, but be prepared to recognise and credit unexpected approaches where they show relevance. Using a ‘best-fit’ approach based on the science content of the answer, first decide which set of level descriptors, Level 1, Level 2 or Level 3, best describes the overall quality of the answer using the guidelines described in the level descriptors in the mark scheme. Once the level is located, award the higher or lower mark. The higher mark should be awarded where the level descriptor has been evidenced and all aspects of the communication statement (in italics) have been met. The lower mark should be awarded where the level descriptor has been evidenced but aspects of the communication statement (in italics) are missing. In summary:
The science content determines the level.
The communication statement determines the mark within a level. Level of response questions on this paper are 3(c) and 5(d)(iv).
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11. Annotations
Annotation Meaning
DO NOT ALLOW Answers which are not worthy of credit
IGNORE Statements which are irrelevant
ALLOW Answers that can be accepted
( ) Words which are not essential to gain credit
__ Underlined words must be present in answer to score a mark
ECF Error carried forward
AW Alternative wording
ORA Or reverse argument
Marking point
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12. Subject-specific Marking Instructions
INTRODUCTION Your first task as an Examiner is to become thoroughly familiar with the material on which the examination depends. This material includes:
the specification, especially the assessment objectives
the question paper
the mark scheme.
You should ensure that you have copies of these materials. You should ensure also that you are familiar with the administrative procedures related to the marking process. These are set out in the OCR booklet Instructions for Examiners. If you are examining for the first time, please read carefully Appendix 5 Introduction to Script Marking: Notes for New Examiners. Please ask for help or guidance whenever you need it. Your first point of contact is your Team Leader.
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Question Answer Marks Guidance
1 (a)
Compound A:
First reaction:
Reagents: HCl/H+ and NaNO2 OR HNO2
Condition: below 5 oC
Second reaction:
Reagents/conditions:
SO3Na
HO
NaOH
5 ALLOW names (including ‘nitrite’, ‘nitrous acid’) ALLOW –O– for –OH ALLOW NaOH as reagent or condition IGNORE temperature for second reaction
(b) –SO3– groups (formed by dissociation of NaSO3)
form ion–dipole bonds with water energy released by bond formation greater than required for bonds broken (AW) (these are) hydrogen bonds between water molecules OR ionic bonds between ions
4
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Question Answer Marks Guidance
(c) FIRST CHECK THE ANSWER ON THE ANSWER LINE Energy change = 266 (kJ mol–1) award 3 marks
∆E for absorption by 1 atom = hc/
∆E for absorption in kJ mol1 = hcNA/1000
= 93
23834
1045010
1002.61000.31063.6
= 266 (kJ mol–1)
3 First and second marks can be scored by correctly substituted figures into the expressions.
(d) 3H2 on top left line and H2 on top right line indication that left-hand side is less than 3 × right-hand side OR that gap between benzene and cyclohexene is less than three times value for cyclohexene
2
(e) (i) FIRST CHECK THE ANSWER ON THE ANSWER LINE
rate =4.35 × 105 award 2 marks
rate =4.35 × 105 AND units = s1 award 3 marks rate doubles with concentration so reaction 1st order to C6H5N2Cl rate = k[C6H5N2Cl]
k = rate / [C6H5N2Cl] = 4.35 × 105 s1
3 Answer can be calculated using either data set.
(ii) only know concentration at start/initial concentration (therefore) collecting larger volumes would not give initial rate
2 IGNORE references to actual percentages
3H2
H2
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Question Answer Marks Guidance
(iii) FIRST CHECK THE ANSWER ON THE ANSWER LINE pH = 1.68 (2 d.p.) award 2 marks n(N2) = 0.050 / 24.0 = 0.00208 mol
(v) two missing values: 3.36 × 10–3 and –8.52 axes drawn, labels correct
x-axis: 1/T / K1 y-axis: ln k scales that use over half of each axis all points plotted correctly with best straight line drawn through points measurement of slope (1.33 ± 0.05 × 104) multiplication by 8.314, division by 103 and change of sign
to give Ea in kJ units [+111 (kJ mol1)]
5 units of 1/T can be missing ALLOW no sign but not minus sign for last mark ALLOW 2 or more sig figs ALLOW any answer rounding to 106–115 (calculated correctly from slope)
Total 27
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Question Answer Marks Guidance
2 (a) (i) CO2 + Ca(OH)2 CaCO3 + H2O
1 IGNORE state symbols
(ii) FIRST CHECK THE ANSWER ON THE ANSWER LINE mass = 0.0625 award 1 mark mass = 0.063 g to 2 sig figs award 2 marks moles CO2 = 15/24000 OR 6.25 x 10–4 mass CaCO3 = (6.25 x 10-4 × 100) = 0.063 g to 2 sig figs
2 ALLOW ECF from first marking point
(b) (i) CO32– + H2O HCO3
– + OH– OR
CO32– + H2O CO2 + 2OH–
1 ALLOW equilibrium signs IGNORE state symbols
(ii) CaCO3 is less soluble/weaker base than Ca(OH)2 so fewer OH–(aq) ions are present (as Ca(OH)2 reacts / CaCO3 forms)
2
(c) (i) FIRST CHECK THE ANSWER ON THE ANSWER LINE pH = 11.6 award 3 marks
[OH–] = 2 × 0.002 = 0.004 (mol dm3)
[H+] = (1 x 10–14/4 x 10–3 =) 2.5 x 10–12 (mol dm3) pH = 11.6
3 ALLOW ECF on third marking point from a given [H+], provided pH is between 10 and 13
(ii) [H+] = 2.1 x 10–5 (mol dm–3) pH = 4.7
2
(iii) contribution of [H+] from water 1 × 10–7 insignificant compared to (AW) 2 × 10–5
2
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Question Answer Marks Guidance
(d) (i) O2 + 2H2O + 4e– ⇌ 4OH–
Fe ⇌ Fe2+ + 2e–
2 ALLOW ‘e’ without minus IGNORE state symbols
ALLOW ½ or ¼ equation for OH
(ii) high [OH–] pushes first equilibrium to left (AW) electrons (formed) push second equilibrium to left so less Fe2+ (and hence rust) formed
(ii) CO2 from air is reacting with water to form H+ turning (the structure of phenolphthalein) from E to D
2
Total 21
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Question Answer Marks Guidance
3 (a) (i) C5H8O2 + 6O2 5CO2 + 4H2O Correct molecular formula of GMV Balanced equation (with ECF)
2 IGNORE state symbols
(ii) FIRST CHECK THE ANSWER ON THE ANSWER LINE ratio = 1 : 1.4 or more sig figs (1.395….) award 3 marks mass CO2 per mole GMV = 5 × 44 = 220 g mass CO2 per mole hexane = 6 × 44 = 264 g Mr values GMV 100; hexane 86 ratio (220/100) : (264/86) = 1 : 1.40
3 OR moles CO2 per mole fuel ALLOW ECF from first marking point
(iii) Comparison of CO2 is important as it is a greenhouse gas/contributes to global warming Should be CO2 per kJ released on burning (AW)
2 ALLOW any unambiguous representation of structure
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Question Answer Marks Guidance
(c)* Please refer to the marking instructions on page 4 of this mark scheme for guidance on how to mark this question. Level 3 (5–6 marks) Candidate identifies the structure of the compound correctly, the evidence for this identification uses information from composition by mass AND IR spectrum AND proton NMR spectrum. The identification is clear and logically structured. The evidence selected is relevant and fully supports the identification. Level 2 (3–4 marks) Candidate identifies the compound as a keto-acid, the evidence for this identification uses information from composition by mass AND EITHER IR spectrum (both points) OR proton NMR spectrum OR using information from IR spectrum AND proton NMR spectrum. The identification has a logical structure. The evidence selected is in the most-part relevant and supports the identification. Level 1 (1–2 marks) Candidate makes an attempt at identifying the compound and the evidence to support this identification uses EITHER information from composition by mass OR IR spectrum OR proton NMR spectrum. There is an attempt at a logical structure with a line of reasoning. The information is in the most part relevant.
6 Indicative scientific points may include: Compound is CH3COCH2CH2COOH Composition by mass:
C5H8O3 calculated. IR spectrum:
COOH from IR at 3200–3600
(• no OH at 3200–3600/3640).
Proton NMR: (any 3)
4 proton environments
peak at 10.5 indicates COOH
peaks at 2.2–2.7 indicate (three) Cs next to (a single) C=O
singlet at 2.2 indicates C with no Hs on adjacent carbons AND triplets at 2.6 and 2.8 indicate Cs with one adjacent CH2.
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Question Answer Marks Guidance
0 marks No response or no response worthy of credit.
Total 15
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Question Answer Marks Guidance
4 (a)
Structure Bond name
2 IGNORE 2+ ALLOW a flat structure, i.e. viewed from above
(b) (i) intensity of colour OR absorbance of (suitable wavelength) light gives a measure of concentration (of complex/ammonia) OR (directly) proportional to concentration (of complex/ammonia)
2
(ii) =
43
2
243
]NH][[Cu
]])[[Cu(NH
1
(iii) 4)44.0)(1.0( xx
x
correct apart from a wrong value for [NH3] (0.4 – 4x) for [NH3]
2 ALLOW ECF from (ii)
(iv) Inaccurate / second student correct because Reaction of (OH– in) ammonia solution with H2SO4 causes equilibrium to move so all the ammonia is neutralised (AW) OR equilibrium 4.1 shifts to the left as the acid reacts with the ammonia so titre / measured concentration will be too large
2
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Question Answer Marks Guidance
(c) Cu(C2H8N2)22+
4
2 ALLOW a more structured formula for ethane diamine 2+ must be present ALLOW Cu(C2H8N2)3
2+ 6
(d) (i) Cu2+ + 2I– CuI + ½ I2 OR doubled
1 IGNORE state symbols
(ii) copper(I) iodide
1 ALLOW a gap between ‘copper’ and ‘(I)’
Total 13
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Question Answer Marks Guidance
5 (a) (i)
BrBr
OH
1 ALLOW any unambiguous formula clearly displaying the structure of the compound.
(ii) 11
1
(iii) two (because ambrinol has two) asymmetric carbon atoms / carbon atoms surrounded by 4 different groups
2
(iv) yes, because same bonds hence same vibrations
2 ALLOW: no, based on analogy with named compound (e.g. carvone) for which enantiomers smell different
(b) vanillin and guaiacol are phenols purple colour with FeCl3
benzaldehyde and vanillin are aldehydes acid dichromate goes green OR Tollens’ reagent OR AgNO3 and NH3 ... give silver (mirror) OR Fehlings solution gives red precipitate
4
(c) IR spectrum that shows how bonds vibrate Shows similarity between spectra of vanillin and combination of guaiacol and benzaldehyde (AW) C–H in arenes
3
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Question Answer Marks Guidance
(d) (i) (atoms with) same atomic number AND different mass number 1 and 7
2 must mention both terms
(ii) 12.01
1
(iii) O
all hydrogen atoms replaced with deuterium atoms
2 Correct skeletal formula required. ALLOW ‘H’/’proton’ for ‘hydrogen atom’ and ‘D’ for deuterium atom To score, answer must state that all the atoms are replaced.
(iv)* Please refer to the marking instructions on page 4 of this mark scheme for guidance on how to mark this question. Level 3 (5–6 marks) Supporting evidence for Turin’s theory AND evidence to discredit the lock and key theory have been correctly identified. There is a clear discussion related to both theories. The evidence selected is relevant and substantiated. The discussion shows a well-developed line of reasoning for the choices of evidence, which is clear and logically structured. Candidate demonstrates a clear and confident knowledge of relevant technical language (names of compounds, ‘deuterated’, ‘vibrate’).
6 Indicative scientific points may include: Statements to support Turin’s theory
deuterated acetophenone smells different (from non-deuterated form)
deuterated benzaldehyde smells different (from non-deuterated form)
because H and D have different masses and hence C–D and C–H vibrate differently
C12 and C13 exchange does not affect smell, since masses only differ slightly
structures of deuterated acetophenone and benzaldehyde given.
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Question Answer Marks Guidance
Level 2 (3–4 marks) Supporting evidence for Turin’s theory AND evidence to discredit lock and key theory have been correctly identified although there is a limited attempt to discuss them. The evidence selected is in the most-part relevant. The discussion has some structure but is limited in scope. Candidate demonstrates a sound grasp of technical language (one spelling error). Level 1 (1–2 marks) Evidence for Turin’s theory has been identified OR Evidence against lock and key theory There is an attempt at a logical structure with a line of reasoning. The information is in the most part relevant. Candidate demonstrates a basic grasp of relevant technical language (several spelling errors). 0 marks No response or no response worthy of credit.
Statements to discredit Lock and Key
since structures are similar, lock and key theory would not predict difference (AW).
structurally related molecules can smell utterly different.
Total 24
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Summary of updates
Date Version Change
January 2019 2.0 Minor accessibility changes to the paper: i) Additional answer lines linked to Level of Response questions ii) One addition to the rubric clarifying the general rule that working should be shown for any calculation questions